Shintani's Zeta Function Is Not a Finite Sum of Euler Products

Home , Dirichlet series

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that “an analogue of [these formulas] for the space of binary cubic forms is currently unknown; although, such a formula would be immensely interesting.” Wright is referring to the Shintani zeta function associated to the space of binary cubic forms. This zeta function was introduced by Shintani [16] and further studied by Datskovsky and Wright [21, 6, 7] and many others. This zeta function is deﬁned as follows: The lattice of integral binary cubic forms is deﬁned by (1.5) V := au3 + bu2v + cuv2 + dv3 : a,b,c,d Z , Z { ∈ } and the discriminant of such a form is given by the usual equation (1.6) Disc(f)= b2c2 4ac3 4b3d 27a2d2 + 18abcd. − − − There is a natural action of GL2(Z) (and also of SL2(Z)) on VZ, given by 1 (1.7) (γ f)(u, v)= f((u, v) γ). · det γ · The Shintani zeta functions are given by the Dirichlet series

1 s (1.8) ξ±(s) := Disc(x) − , Stab(x) | | x SL2(Z) VZ | | ∈Disc(Xx)\>0 ± and Shintani proved that they have analytic continuation to C and satisfy a functional equation. These zeta functions are interesting for a variety of reasons; see, e.g., [18] and [7] for applications to counting cubic extensions of number ﬁelds. Also note that Datskovsky and Wright’s work yields an adelic formulation of Shintani’s zeta function, similar to (1.4). Some further motivation for Wright’s question is provided by work of Ibukiyama and Saito [10]. They consider the zeta functions associated to the prehomogeneous vector spaces of n n symmetric matrices, for n > 3 odd. In particular they prove, among many other interesting× results, explicit formulas for these zeta functions as sums of two products of the Riemann zeta function. With these results in mind, one might hope to prove similar such formulas for the zeta functions (1.8). In this note we answer Wright’s question in the negative, and prove that no such formulas exist.

Theorem 1.1. Neither of the Shintani zeta functions ξ±(s) deﬁned in (1.8) admits a representation as a ﬁnite sum of Euler products.

s k In other words, if we write ξ±(s)= n a(n)n− , we cannot write a(n)= i=1 cibi(n) for real numbers ci and multiplicative functions bi(n). It is interesting to note that these zetaP functions do have representationsP as inﬁnite sums s of Euler products. Of course, there is a tautological such representation, writing n− = s (1+n− ) 1 and thus regarding each term in (1.8) as a diﬀerence of Euler products. However, Datskovsky− and Wright [6] proved the much more interesting formula

2 s Rk(2s) (1.9) ξ±(s)= ζ(4s)ζ(6s 1) Disc(k) − , − o(k)| | Rk(4s) Xk SHINTANI’S ZETA FUNCTION IS NOT A FINITE SUM OF EULER PRODUCTS 3

where the sum is over all number ﬁelds of degree 1, 2, or 3 (up to isomorphism) of the correct sign; o(k) is equal to 6, 2, 1, or 3 when k is trivial, quadratic, cubic and non-Galois, or cubic 3 and cyclic respectively; and Rk(s) is equal to ζ(s) ,ζ(s)ζk(s),ζk(s),ζk(s) respectively. The proof of Theorem 1.1 is not diﬃcult, and it illustrates an application of recent work of Cohen and Morra ([4]; see also Morra’s thesis [11] for a longer version with more examples). For a fundamental discriminant D, they prove explicit formula for the Dirichlet series

2 s (1.10) ΦD(s) := a(Dn )n− , n X where a(Dn2) is the number of cubic fields of discriminant Dn2. More generally, they prove a formula for the Dirichlet series counting cubic extensions of number fields K/k, such that the normal closure of K contains a fixed quadratic subextension K2/k. The formula is rather complicated, and it involves sums over characters of the 3-part of ray class groups associated to Q(√D, √ 3) (or more generally to K2(√ 3)). However, these formulas take much simpler forms when− we assume k = Q, and when− we further restrict to D for which we can control these class groups. We will apply the following special case of their result: Theorem 1.2 (Cohen-Morra [4]). If D < 0, D 3 (mod 9), and 3 ∤ h(D), then we have ≡ 2 s 1 1 2 2 − (1.11) a(Dn )n = + 1+ s 1+ s . −2 2 3 −3 p n D =1 X ( Yp ) Furthermore, if D > 0, D 3 (mod 9), and 3 ∤ h( D/3), then there exists a cubic field E of discriminant 27D such≡ that − −

2 s 1 1 2 2 − (1.12) a(Dn )n = + 1+ s 1+ s −2 6 3 −3 p n D =1 X ( Yp ) 1 1 ωD(p) + 1 s 1+ s , 3 − 3 −3 p D =1 ( Yp ) where ω (p)=2 if p splits completely in E, and ω (p)= 1 otherwise. D D − Remark. In fact, p splits completely in E if and only if it does so in its Galois closure E(√ 3D), and this version of the condition will be more convenient in our proof. − The equation (1.11) is Corollary 7.9 of [4]. The analogue (1.12) does not appear in [4, 11], so in Section 3 we describe how this formula follows from Cohen and Morra’s work. Moreover, in forthcoming work with Cohen [5] we will generalize (1.12) to any D, positive or negative, and we will also prove an analogous formula for quartic fields having fixed cubic resolvent. We can now summarize our proof of Theorem 1.1, in the easier negative discriminant s case. First observe that n a(n)n− := ξ−(s) ζ(s) cannot be an Euler product: By (1.11), we may find a negative D 3 (mod 9) and− many primes p for which a( D) = 0 and a( Dp2) > 0. P ≡ − − 4 FRANK THORNE

To prove that that ξ−(s) ζ(s) is not a sum of k Euler products, for any k > 1, we choose appropriate discriminants D− and primes p such that a( D p2) > 0 if and only if i = j. i i − i j Then a(n) “takes too many parameters to describe,” and we will formalize this using an elementary linear algebra argument. The argument for the positive discriminant case is mildly more complicated, but essentially the same.

Closing remarks. Although we cheerfully admit that a positive result would be more interesting than the present paper, we submit that our negative result is interesting as well. One motivation for our work comes from our previous paper [20], where we studied the Shintani zeta functions analytically, obtaining (limited) results on the location of the zeroes. These zeta functions essentially fit into the Selberg class, which naturally leads to a host of questions. For example, do most of the zeroes of the Shintani zeta functions lie on the critical line? Work of Bombieri and Hejhal [3] establishes that this is true for certain finite sums of L-functions, conditional on standard hypotheses for these L-functions (including GRH). As Shintani’s zeta functions lie outside the scope of [3], even a good conjecture would be extremely interesting. Our work was also motivated by the desire to incorporate recent developments in the study of cubic fields into the study of the Shintani zeta function. In addition to the work of Cohen and Morra, we point out the work of Bhargava, Shankar, and Tsimerman [2], Hough [9], and Zhao [24] for further approaches to related questions. We write this paper in optimism that further connections will be found among the various approaches to the subject.

2. Acknowledgments I would like to thank Takashi Taniguchi, Simon Rubinstein-Salzedo, and the referee for useful comments, and Yasuo Ohno for pointing out the reference [10] to me, which indirectly motivated this paper. Finally, I would like to thank Henri Cohen and Anna Morra for answering my many questions about their interesting work.

3. Proof of Theorem 1.2 In this section, we briefly describe the work of Cohen and Morra [4, 11], and discuss how (1.12) follows from their more general results.1 In a followup paper with Cohen [5], we will extend Theorem 1.2 and the proof given here to cover the case where D is any fundamental discriminant. The original work of Cohen and Morra is more general still, enumerating relative cubic extensions of any number field. The new ingredient here (and in [5]) is that the Cohen-Morra formulas are somewhat abstract, and it is not obvious that they yield explicit formulas like (1.12). Suppose then that K/Q is a cubic field of discriminant Dn2, where D 1, 3 is a fundamental discriminant, and let N be the Galois closure of K. Then N(√6∈3) { is− a} cyclic − cubic extension of L := Q(√D, √ 3), and Kummer theory implies that N(√ 3) = L(α1/3) − − 1Our specific references are to the published paper [4], but we also recommend [11] for its enlightening additional explanation and examples. SHINTANI’S ZETA FUNCTION IS NOT A FINITE SUM OF EULER PRODUCTS 5

for some α L. We write (following [4, Remark 2.2]) ∈ (3.1) Gal(L/Q)= 1, τ, τ , ττ , { 2 2} where τ, τ2, ττ2 fix √D, √ 3, √ 3D respectively. The starting point of [4] is− a correspondence− between such fields K and such elements α. In particular, isomorphism classes of such K are in bijection with equivalence classes of elements 3 3 1 = α L×/(L×) , with α identified with its inverse, such that ατ ′(α) (L×) for τ ′ 6 ∈ 3 ∈ ∈ τ, τ . We say (as in [4, Definition 2.3]) that α (L×/(L×) )[T ], where T F [Gal(L/Q)] { 2} ∈ ⊆ 3 is defined by T = τ +1, τ2 +1 , and the notation [T ] means that α is annihilated by T . This bijection opens the{ door to Cohen-Morra’s} further study of such α in terms of the ideals of L. Assume that D 3 (mod 9), placing us in case (5) of [4, p. 464]. The formula (1.11) is Corollary 7.9 of≡ [4], and so we assume that D > 0, for which the conditions for (1.12) 3 are discussed in Remark 7.8. Define Gb, as in [4, Theorem 6.1], to be Clb(L)/Clb(L) [T ], where Clb(L) is the ray class group of an ideal b. By Remark 7.8, Gb is of order 1 (for all b considered in Theorem 6.1), except when b = (3√ 3), one case where it has order 3. In this setting, the main theorem (Theorem 6.1− of [4]) reduces to a formula of the shape

2 s 1 1 ωχ(p) − (3.2) a(Dn )n = + Ab(s) ωχ(3) 1+ s , −2 2 c p n b χ Gb −3D X X∈B X∈ Yp =1 where for p = 3 we have 6 2 if χ(pc)= χ(pτ(c)) , ω (p)= χ 1 if χ(pc) = χ(pτ(c)) . (− 6 We will define c later, but we leave the definitions of the other quantities to [4, 11] (see also our forthcoming followup [5]). Most of the computations needed to verify (1.12) are either completely straightforward or carried out in [4, 11]. In particular, the contributions of the trivial characters are computed there. The one remaining step requiring a substantial argument is to prove that for each nontrivial character χ of G(3√ 3), we have ωχ(p)= ωD(p). This involves a bit of class field theory, and we give the proof here.− 2 3 Define Gb′ := Clb(L)/Clb(L) , so that Gb′ is a 3-torsion group containing Gb. We have a canonical decomposition of Gb′ into four eigenspaces for the actions of τ and τ2, and we write

(3.3) G′ Gb G′′ b ≃ × b where Gb′′ is the direct sum of the three eigenspaces other than Gb. Note that Gb′′ will contain the classes of all principal ideals generated by rational integers coprime to 3; any such class 3 in the kernel of T will necessarily be in Clb(L) . We may thereby write Gb Clb(L)/H, where H is of index 3 and contains the classes of all principal ideals coprime to≃ 3. By class ﬁeld theory, there is a unique abelian extension E1/L for which the Artin map induces an isomorphism Gb Gal(E /L). It must be cyclic cubic, as Gb is, and the unique- ≃ 1 ness forces E1/L to be Galois over Q, as the group Gb is preserved by τ and τ2 and hence

2 ′ ′′ We have followed the notation of [4] where practical, but the notations Gb, Gb, E, E1,ωD(p), and some of the notation appearing in (3.2) are used for the ﬁrst time here and do not appear in [4]. 6 FRANK THORNE

all of Gal(L/Q). We have Gal(E /Q) S C : τ and τ Gal(L/Q) both act nontrivially 1 ≃ 3 × 2 2 ∈ on Gb, and under the Artin map this implies that τ, τ2 both act nontrivially on Gal(E1/L) by conjugation. This forces ττ to commute3 with Gal(E /L). As ττ fixes Q(√ 3D), this 2 1 2 − implies that E contains a cubic extension E/Q with quadratic resolvent Q(√ 3D), which 1 − is unique up to isomorphism. Any prime p which splits in Q(√ 3D) must either be inert or totally split in E. − 3D For each prime p with − = 1, write p = cτ ′(c), where τ ′ is the element τ or τ of p OL 2 Gal(L/Q) described previously. Here c is either prime or a product of two primes pττ2(p) D depending on whether is 1 or 1. For each such p, ω (p) = 2 if χ(pc) = χ(pτ ′(c)), p − χ and ω (p) = 1 if χ(pc) = χ(pτ ′(c)). With the isomorphism Gb Clb(L)/H, we have χ − 6 ≃ χ((p)) = 1, so that χ(c) is well defined and equal to χ(pc) (and similarly for χ(τ ′(c))). Therefore, the computation χ(c)χ(τ ′(c)) = χ((p)) = 1 implies that χ(c) and χ(τ ′(c)) are complex conjugates, and thus ωχ(p) is 2 or 1 depending on whether χ(c) = 1 or not. We claim that that χ(c) = 1 if and only− if p splits completely in E. Suppose first that c is prime in . Then χ(c) = 1 if and only if c splits completely in E /L, in which case (p) OL 1 splits into six ideals in E1, which happens if and only if p splits completely in E. Suppose now that c = pττ2(p) in L. We claim that χ(c) = 1 if and only if χ(p) = 1; this follows as p and ττ2(p) have the same Frobenius element in E1/L (which in turn is true because they represent the same element of Gb), and χ(c)= χ(pττ2(p)) = 1. Now χ(p)=1 if and only if p splits completely in E1/L, in which case (p) splits into twelve ideals in E1; for this it is necessary and sufficient that p split completely in E. This proves that ωχ(p)= ωD(p), as desired, provided that we verify that the discriminant of E is as claimed. As its quadratic resolvent is Q(√ 3D), we observe that Disc(E) = r2( D/3) for some integer r divisible only by 3 and prime− divisors of D. No prime ℓ > 3 can− divide r, because ℓ3 cannot divide the discriminant of any cubic field. Similarly 2 cannot divide r, as if 2 D then 4 D, but 16 cannot divide the discriminant of a cubic field. | | Therefore r must be a power of 3. We cannot have r = 1, as E(√ 3D)/Q(√ 3D) would be an unramified cubic extension, but h( D/3) = 1. r cannot be 3,− nor can we− have 27 r, as the 3-adic valuation of a cubic field discriminant− is never 2 or larger than 5. By process| of elimination, r = 9. Remark. Belabas [1] has computed a table of all cubic fields K with Disc(K) < 106, and we used his table and PARI/GP [15] to double-check (1.12). | |

4. Proof of Theorem 1.1 We will require the following lemma, which we extract from a generalization of the Davenport-Heilbronn theorem [8]. Lemma 4.1. Given any residue class a (mod 36m) with a 21 (mod 36) or a 5 (mod 12), and (6a, m)=1, there exist inﬁnitely many negative fundamental≡ discriminants≡n a (mod 36m) which are not discriminants of cubic ﬁelds. ≡

3 More properly speaking, we choose an arbitrary lift of ττ2 to Gal(E1/Q) and denote it also by ττ2; it is this lift which commutes with Gal(E1/L). SHINTANI’S ZETA FUNCTION IS NOT A FINITE SUM OF EULER PRODUCTS 7

Proof. This follows from quantitative versions of the Davenport-Heilbronn theorem in arithmetic progressions. See [18] for sharp quantitative results; alternatively, the methods of [13] may be adapted to give an easier proof. In particular, the results stated in [18] imply that the average number of cubic fields K 1 per negative fundamental discriminant is equal to 2 , and that this average is the same when restricted to any arithmetic progression as above containing fundamental discriminants. The lemma follows from the fact that this average is less than 1. Proof of Theorem 1.1. For simplicity we begin by proving the theorem for the negative discriminant zeta function ξ−(s). The basic idea is the same for all cases. For reasons that will become apparent later, we begin by subtracting the Riemann zeta function ζ(s). Suppose that the function ξ−(s) ζ(s) has a representation − k s s (4.1) ξ−(s) ζ(s)= a(n)n− = c b (n) n− , − i i n n i=1 X X X where the ci are nonzero real numbers and the bi(n) are multiplicative functions. We will obtain a contradiction, implying Theorem 1.1. Begin by using Lemma 4.1 to choose k odd negative fundamental discriminants ni 21 (mod 36), coprime to each other apart from the common factor of 3, for which there are≡ no cubic fields of discriminant ni. 3ni ni Now, for 1 i k, choose primes pi 1 (mod 3) for which − = = 1 for i = j, ≤ ≤ ≡ pj pj − 6 3ni ni and − = = 1 for each i. (These conditions amount to arithmetic progressions mod pi pi n .) We will count the number of cubic rings with discriminant n , n p2, and n p2, for each i i i i i j i and j. Q For each ni, there is exactly one quadratic field of discriminant ni. There is exactly one 2 nonmaximal reducible ring of discriminant nipj for j = i, and three such rings of discriminant 2 6 nipi ; these nonmaximal rings are counted by (1.9). By hypothesis, there are no cubic fields of discriminant ni. By (1.11), there are no cubic 2 fields (or nonmaximal rings) of discriminant nipj for j = i, and there is one cubic field of 2 6 discriminant nipi . As the Shintani zeta function counts noncyclic cubic fields with weight 2, and quadratic rings with weight 1, we conclude that a(n p2) a(n ) is equal to zero if i = j, and is positive i j − i 6 if i = j. This fact is enough to contradict (4.1). To see this, let B be the k k matrix with × (r, s)-entry csbs(nr), where cs and bs(n) are as in (4.1), and let Ci be the vector (e.g., the 2 n 1 matrix) whose rth row is equal to br(pi ) br(1). Then BCi is the vector whose rth row× is equal to − k k (4.2) c b (n ) b (p2) b (1) = c b (n p2) c b (n ) = a(n p2) a(n ). ℓ ℓ r ℓ i − ℓ ℓ ℓ r i − ℓ ℓ r r i − r ℓ=1 ℓ=1 X X This is equal to zero if and only if i = r, and hence the column matrices BCi are linearly independent over R, and so B is invertible.6 Now write C′ for the vector consisting of k ones, so that the rth row of BC′ is equal to a(nr). We chose nr so that there are no cubic fields of discriminant nr for any r, so that the 8 FRANK THORNE

Shintani zeta function counts only the quadratic ﬁeld of discriminant nr. By (4.1), a(nr)=0 for each nr, and hence BC′ = 0. But this contradicts the invertibility of B; therefore the representation (4.1) cannot exist.

Positive discriminants. To prove our result for the positive Shintani zeta function, we again define a(n) as in (4.1), again subtracting ζ(s). We choose positive fundamental discriminants ni 21 (mod 36) as before, with no other prime factors in common, and such that there are no≡ cubic fields of discriminant n /3. It follows (by (1.12), or more simply by the Scholz − i reflection principle) that there are also no cubic fields of discriminant ni. We choose primes pi as before, only this time we require that ωni (pi) = 2 so that there will again be a cubic field of discriminant n p2. This condition holds if p splits completely in E(√ 3n ), where E i i i − i is as in (1.12). Roughly speaking, we can find such a pi because this condition is non-abelian. 3ni More precisely, we want to choose p so that − = 1 (implied if p splits completely i pi i ni in E(√ 3ni)), that = 1 (implied by the previous statement when p 1 (mod 3) and pi − 3 nj ≡ therefore − = 1), and that )= 1 for j = i. As the extensions E(√ 3ni), Q(√ 3), pi pi − 6 − − and Q( n )(j = i) are all Galois and disjoint over Q, the Galois group of their compositum √ j is the direct product6 of their Galois groups, and thus the Chebotarev density theorem implies that these splitting conditions may all be simultaneously satisfied. Finally, note that the quadratic fields and rings may be handled exactly as in the negative discriminant case, and so the remainder of the proof is unchanged.

4.1. Generalizations of our results. Our proof generalizes with minimal modiﬁcation to some additional Dirichlet series related to Shintani’s zeta functions. We now describe examples of such Dirichlet series and the modiﬁcations required.

The Dirichlet series for cubic ﬁelds. A Dirichlet series related to the Shintani zeta function is that simply counting cubic ﬁelds:

s (4.3) F ±(s) := Disc(K) − . | | [K:Q]=3 Disc(XK)>0 ± This does not seem to have any nice analytic properties (such as meromorphic continuation to C), but we cannot rule such properties out. Needless to say, if F ±(s) had such analytic properties then this would have interesting consequences for the distribution of cubic fields. For example, this would presumably yield another proof of the Davenport-Heilbronn theorem, possibly with better error terms than are currently known. Our argument also proves that neither series F ±(s) can be represented as a finite sum of Euler products. The proof is exactly as before, only easier because there are no quadratic s fields to count. We write n a(n)n− = F ±(s) instead of subtracting ζ(s) as in (4.1); we 2 2 thus have a(ni) = a(nipj ) = 0 for each ni when j = i, and a(nipi ) > 0, and hence the conclusion of the argumentP works. 6 1 The same argument also works if each field is counted with the weight Aut(K) ; indeed, our proof does not see any cyclic cubic fields. | | SHINTANI’S ZETA FUNCTION IS NOT A FINITE SUM OF EULER PRODUCTS 9

+ Linear combinations. Suppose that ξ(s) = C1ξ (s)+ C2ξ−(s) is a linear combination of the usual Shintani zeta functions, with C C = 0. Of particular interest are the cases 1 · 2 6 C1 = √3, C2 = 1, as the resulting zeta functions have particularly nice functional equa- tions [6, 12, 14]. We± can prove once again that ξ(s) is not a ﬁnite sum of Euler products: We choose negative fundamental discriminants ni as above, and the same proof works exactly. (Instead of subtracting ζ(s) we instead subtract C2ζ(s).) The point is that all of the n considered in the proof are 3 (mod 4), corresponding to ﬁelds and nonmaximal rings of discriminant n, and so no≡ positive fundamental discriminants enter our calculations. We could similarly− work only with positive discriminants and ignore the negative ones.

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